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1 Electromagnetism Physics 15bLecture #4 Divergence and Laplacian Purcell 2.7–2.12 What We Did Last Time Used Gauss’s Law on infinite sheet of charge Uniform electric field E= 2πσabove and below the sheet Electric field has energy with volume density given by Defined electric potential by line integral Electric field is negative gradient of electric potential Potential due to charge distribution: or Total energy of a charge distribution: u=E28πφ21=−E⋅dsP1P2∫=φ(P2)−φ(P1)unit: erg/esu = statvolt E=−∇φφ=qjrjj=1N∑φ=dqr∫U=12ρφdv∫
2 Today’s Goals Introduce divergence of vector field How much “flow” is coming out per unit volume Translate Gauss’s Law into a differential (local) form Gauss’s Divergence Theorem connects the two forms Look in the energy again Equivalence of and Define the Laplacian = divergence of gradient Re-express Gauss’s Law with a Laplacian Study mathematical properties of Laplace’s equation Conclude with a Uniqueness Theorem U=12ρφdv∫U=E28πdV∫Shrinking Gauss’s Law Charge is distributed with a volume density ρ(r) Draw a surface Senclosing a volume VGuass’s Law: Now, make V so small that ρis constant inside V As we make Vsmaller, the total flux out of Sscales with V Therefore: LHS is “how much E is flowing out per unit volume” Let’s call it the divergence of EE⋅daS∫=4πρdvV∫Total charge in VE⋅daS∫=4πρVfor very small V limV→0E⋅daS∫V=4πρ
3 Divergence In the small-Vlimit, the integral depend on volume, but not on the shape We can use a rectangular box Consider the left (S1) and right (S2) walls Add up all walls: divE≡limV→0E⋅daS∫V=4πρdx dy dz E(x+dx,y,z)E(x,y,z)S1 S2 E⋅daS1∫=E(x,y,z)⋅(−ˆx)dydzE⋅daS2∫=E(x+dx,y,z)⋅ˆxdydzSum=Ex(x+dx)−Ex(x)()dydz=∂Ex∂xdxdydzE⋅daS∫=∂Ex∂x+∂Ey∂y+∂Ez∂z⎛⎝⎜⎞⎠⎟V=∇ ⋅E()Vdiv E Gauss’s Law, Local Version We now have Gauss’s Law for a very small volume/surface Connects local properties of Ewith the local charge density Divergence has the easy form in Cartesian coordinates